Can you construct 5x5 and 6x6 “completely-odd” matrices?

Question

An n x n matrix is said to be completely-odd if:

the numbers in the matrix are either 0 or 1

and

for every number in the matrix, the sum of the numbers in its neighborhood is odd where the neighborhood of a number includes the number itself and its horizontally and vertically adjacent numbers.

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EXAMPLE:

Shown below are three views of the same 4x4 completely-odd matrix:

The number in row 1 column 1 with its neighborhood highlighted has a neighborhood sum of 1+0+0=1 which is odd:

The number in row 1 column 2 with its neighborhood highlighted has a neighborhood sum of 1+0+1+1=3 which is odd:

The number in row 3 column 2 with its neighborhood highlighted has a neighborhood sum of 1+1+1+1+1=5 which is odd:

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PUZZLES:

Can you construct a 5x5 completely-odd matrix?
Can you construct a 6x6 completely-odd matrix?

Answer 1

$n=6$:

$$\begin{matrix}1 & 0 & 1 & 1 & 0 & 1\\0 & 1 & 1 & 1 & 1 & 0\\1 & 1 & 1 & 1 & 1 & 1\\1 & 1 & 1 & 1 & 1 & 1\\0 & 1 & 1 & 1 & 1 & 0\\1 & 0 & 1 & 1 & 0 & 1\end{matrix}$$

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Just wanted to add a bit more about my thoughts on how to construct a solution for any $n$:

First, I want to assume that the matrix is symmetric. This is certainly true in most of the answers here, and my goal isn't to construct all possible solutions anyway. With this assumption, it is clear that all the entries in the diagonal must be $1$. By the symmetry, we can focus only on the lower triangle. Let the entries in the diagonal below the principal diagonal be $a_1,\ldots,a_{n-1}$. The lower triangle of the matrix looks like this now $$\begin{matrix} 1 \\ a_1 & 1 \\ & a_2 & 1 \\ & & \ddots & \ddots \\ & & & a_{n-1} & 1\end{matrix}$$ Everything after this point is completely determined in terms of $a_i$'s by the odd-sum constraints. For instance, the entry below $a_1$ must be $1 + a_1 \pmod 2$ so that the sum of all neighbours of $a_1$ (including itself) is $1 \pmod 2$. However, $a_i$'s are not all arbitrary. To fill up the entire (lower triangle of the) matrix, we use only the odd-sum constraints up to $n-1$th row. The odd-sum constraints of the $n$th row give $n-1$ equations in $a_i$'s. I don't yet have a proof that these $n-1$ equations have a solution for any $n$, but for $n\le 6$, it gives precisely the answers here (up to reflections/rotations of the matrix).

Answer 2

$5\times5$

\begin{matrix}0&0&0&1&1\\1&1&0&1&1\\1&1&1&0&0\\0&1&1&1&0\\1&0&1&1&0\end{matrix}

Answer 3

For the 6x6 case,

we can reuse some corners from the 4x4:

Answer 4

$5 \times 5$:

AAOBB
AOOOB
OOOOO
COOOD
CCODD

AAEBB
AEEEB
COEOD
CCFDD
CFFFD

The Os can be vertically aligned instead if the diagram is rotated by 90 degrees, so the sum of the middle square's neighbors is even, making this square's value 1. Also, the sum of the squares from A to F is even because there are 6 groups.

AACBB
ACCCB
GO1OG
DFFFE
DDFEE

The sum of the groups from A to F is also even but 1 is odd, which means the sum of the G cells is odd (opposites).

A B X C D
E F G H I
Z J 1 K 1-Z
L M N O P
Q R 1-X S T

A+B+E = A+B+F+X = 1 (mod 2)
E = F+X (mod 2)

A.........Z+F...X.....H+1-Z D
F+X......F......G.....H........X+H
Z.........J.........1.....K.......1-Z
1+M-X..M......N.....O........1+O-X
Q..........Z+M..1-X 1+O-Z T

G = X+F+H (mod 2)

A.........Z+F...X..........H+1-Z.....D
F+X......F......X+F+H.....H........X+H
Z.......F+Z+M..1.....H+O+1-Z....1-Z
1+M-X..M....M+O+1-X...O...1+O-X
Q..........Z+M..1-X.....1+O-Z.....T

A+Z+X = odd
X+D-Z = X+D+Z = even
D = 1-A (mod 2)

A.........Z+F...X..........H+1-Z.....1-A
F+X......F......X+F+H.....H........X+H
Z.......F+Z+M..1.....H+O+1-Z....1-Z
1+M-X..M....M+O+1-X...O...1+O-X
1-A..........Z+M..1-X.....1+O-Z.....A

5F+2Z+2X+H+M = F+H+M = odd
H+F+O = odd, M= O

A.........Z+F...X....F+1-Z.....1-A
F+X......F......X.....F........X+F
Z..........Z......1....1-Z....1-Z
1+F-X..F......1-X...F......1+F-X
1-A......Z+F...1-X..1+F-Z.....A

F=1

A.........Z+1...X....Z.....1-A
1+X......1......X.....1........X+1
Z..........Z......1....1-Z....1-Z
X...........1......1-X.....1......X
1-A......Z+1...1-X..Z.....A

The only rule left is A+Z+X = 1 (mod 2). Let A=1:

11000
11011
00111
01110
01101

$6 \times 6$: